At $x=\frac{5\pi}{6},\,f(x)=2\sin 3x+3\cos 3x$ is

neither maximum nor minimum

The correct answer is Option (4) → neither maximum nor minimum ##

We have,

$f(x) = 2\sin 3x + 3\cos 3x$

$∴f'(x) = 2 \cdot \cos 3x \cdot 3 + 3(-\sin 3x) \cdot 3$

$\Rightarrow f'(x) = 6 \cos 3x - 9 \sin 3x \quad \dots(i)$

Now,

$f''(x) = -18 \sin 3x - 27 \cos 3x$

$= -9(2 \sin 3x + 3 \cos 3x)$

$∴f'\left(\frac{5\pi}{6}\right) = 6 \cos \left(3 \cdot \frac{5\pi}{6}\right) - 9 \sin \left(3 \cdot \frac{5\pi}{6}\right)$

$= 6 \cos \frac{5\pi}{2} - 9 \sin \frac{5\pi}{2}$

$\left[ ∵\cos(2\pi + \theta) = \cos \theta, \sin(2\pi + \theta) = \sin \theta \right]$

$= 6 \cos \left(2\pi + \frac{\pi}{2}\right) - 9 \sin \left(2\pi + \frac{\pi}{2}\right) = 6 \cos \frac{\pi}{2} - 9 \sin \frac{\pi}{2}$

$= 0 - 9 \neq 0 \quad \left[ ∵\cos \frac{\pi}{2} = 0, \sin \frac{\pi}{2} = 1 \right]$

So, $x = \frac{5\pi}{6}$ cannot be point of maxima or minima.

Hence, $f(x)$ at $x = \frac{5\pi}{6}$ is neither maximum nor minimum.